Let's assume that $X$ is a continuous random variable. Then, around variables like $Y$ there are two possible confusions related to terminology:
$Y$ is a censored version of $X$ (and sometimes people confuse "censoring" with "truncating"). Also, $Y$ has a mixed distribution (and sometimes people confuse "mixed" distributions with "mixture" distributions).
$Y$ is a censored version of $X$ because a) we observe only non-negative values and b) when we observe $0$ we only know that the corresponding realization of $X$ was $x\leq 0$, i.e. we learn only that $X$ fell in an interval, not its specific value. This also creates the mixed distribution, i.e. a distribution that has a discrete part and a continuous part:
At value $0$, non-zero probability mass concentrates, equal to $P(X\leq 0)$. For $X>0$ we have $Y=X$. So the CDF of Y is
$$F_Y(y) = \left\{
\begin{array}{lr}
F_X(y)& : y > 0\\
F_X(0) & : y = 0\\
0 & \text{elsewhere}
\end{array}
\right\}$$
and $\lim_{y\rightarrow \infty}F_Y(y)=1$.
As for its probability -mass function? - density function? -well, see the illuminating and educative comments below on the matter, as well as ways to express it in one row. Let's say we have a point where non-zero probability mass concentrates, $P(Y=0) = P(X \leq 0)<1$, while for $y>0$ we have a continuous function $f_X(y)$ that does satisfy $\int_{0}^{\infty}f_X(y)dy = 1- P(X \leq 0)$ , while this entity is $0$ everywhere else. So in all, it sums up to unity.